Library / Civil / CIV-34
Civil
Joint displacement by virtual work in a truss
Displacement of a joint in the direction of a virtual unit load, from the real member forces N, the member forces n caused by that unit load, and each member’s length, modulus and area.
- Ref
- CIV-34
- Rev
- v1
- Field
- Civil
- Result
- δ [m]
The displacement of a joint in the direction of a unit virtual load is found from δ = ΣNnL/(EA), where N is the real force in each member, n the force in it from the unit load, and L, E and A its length, modulus and area. It is the work done by the virtual forces on the real deformations. Members that are unloaded in either system contribute nothing.
Displacement of a joint in the direction of a virtual unit load, from the real member forces N, the member forces n caused by that unit load, and each member’s length, modulus and area. Tension is positive in both systems. Pin joints, small displacements and axial deformation only. Up to three members here; leave unused ones with N = 0.
Calculator
- Contribution of member 1, N₁n₁L₁/(E₁A₁)3×10−4 m
- Contribution of member 2, N₂n₂L₂/(E₂A₂)1×10−4 m
- Contribution of member 3, N₃n₃L₃/(E₃A₃)0 m
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The joint moves 0.4 mm in the direction of the virtual load; a positive sign means it moves the same way.
- A member whose N and n have the same sign adds to the displacement, and one with opposite signs subtracts from it.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (m) | Local step | Change in result |
|---|---|---|---|---|---|---|
| N1 Actual force in member 1 | −1,000,000 to 1,000,000 N | −0.0149 | 0.0151 | 0.03 | +40000 N | +150 % |
| N2 Actual force in member 2 | −1,000,000 to 1,000,000 N | −0.0097 | 0.0103 | 0.02 | +40000 N | −100 % |
| E1 Modulus of member 1 | 1e10 to 4e11 Pa | 2.5e-4 | 0.0061 | 0.00585 | +10% | −6.82 % |
| n1 Unit-load force in member 1 | −5 to 5 | −0.0014 | 0.0016 | 0.003 | +0.2 | +15 % |
| A1 Area of member 1 | 1e-4 to 0.01 m² | 1.3e-4 | 0.0031 | 0.00297 | +10% | −6.82 % |
| n2 Unit-load force in member 2 | −5 to 5 | −7e-4 | 0.0013 | 0.002 | +0.2 | −10 % |
| L1 Length of member 1 | 0.5 to 20 m | 1.5e-4 | 0.0021 | 0.00195 | +10% | +7.5 % |
| E2 Modulus of member 2 | 1e10 to 4e11 Pa | 3.5e-4 | 0.0023 | 0.00195 | +10% | −2.27 % |
| A2 Area of member 2 | 1e-4 to 0.01 m² | 3.1e-4 | 0.0013 | 9.9e-4 | +10% | −2.27 % |
| L2 Length of member 2 | 0.5 to 20 m | 3.125e-4 | 8e-4 | 4.875e-4 | +10% | +2.5 % |
| N3 Actual force in member 3 | −1,000,000 to 1,000,000 N | 4e-4 | 4e-4 | 0 | +40000 N | 0 % |
| n3 Unit-load force in member 3 | −5 to 5 | 4e-4 | 4e-4 | 0 | +0.2 | 0 % |
| L3 Length of member 3 | 0.5 to 20 m | 4e-4 | 4e-4 | 0 | +10% | 0 % |
| E3 Modulus of member 3 | 1e10 to 4e11 Pa | 4e-4 | 4e-4 | 0 | +10% | 0 % |
| A3 Area of member 3 | 1e-4 to 0.01 m² | 4e-4 | 4e-4 | 0 | +10% | 0 % |
Sweep of Actual force in member 1
| Actual force in member 1 (N) | Joint displacement (m) |
|---|---|
| −1,000,000 | −0.0149 |
| −750,000 | −0.01115 |
| −500,000 | −0.0074 |
| −250,000 | −0.00365 |
| 0 | 1e-4 |
| 250,000 | 0.00385 |
| 500,000 | 0.0076 |
| 750,000 | 0.01135 |
| 1,000,000 | 0.0151 |
Sweep one variable
Variables and how each one contributes
Joint displacement
What the equation gives you. Everything else feeds it.
Actual force in member 1
The real axial force in member 1, in N (tension positive). Scales that member’s contribution in direct proportion. Raising it by 10 % (to 22,000) moves the result by +7.5 %, from 4×10−4 to 4.3×10−4 m.
Actual force in member 2
The real axial force in member 2, in N (tension positive). Scales that member’s contribution in direct proportion. Raising it by 10 % (to −11,000) moves the result by +2.5 %, from 4×10−4 to 4.1×10−4 m.
Actual force in member 3
The real axial force in member 3, in N (tension positive). Scales that member’s contribution in direct proportion. Raising it by 10 % (to 0) moves the result by +0 %, from 4×10−4 to 4×10−4 m.
Unit-load force in member 1
The force in member 1 from the unit virtual load (tension positive). Scales that member’s contribution in direct proportion. Raising it by 10 % (to 1.1) moves the result by +7.5 %, from 4×10−4 to 4.3×10−4 m.
Unit-load force in member 2
The force in member 2 from the unit virtual load (tension positive). Scales that member’s contribution in direct proportion. Raising it by 10 % (to −0.55) moves the result by +2.5 %, from 4×10−4 to 4.1×10−4 m.
Unit-load force in member 3
The force in member 3 from the unit virtual load (tension positive). Scales that member’s contribution in direct proportion. Raising it by 10 % (to 0) moves the result by +0 %, from 4×10−4 to 4×10−4 m.
Length of member 1
The length of member 1, in m. A longer member stretches more under the same force. Raising it by 10 % (to 3.3) moves the result by +7.5 %, from 4×10−4 to 4.3×10−4 m.
Length of member 2
The length of member 2, in m. A longer member stretches more under the same force. Raising it by 10 % (to 4.4) moves the result by +2.5 %, from 4×10−4 to 4.1×10−4 m.
Length of member 3
The length of member 3, in m. A longer member stretches more under the same force. Raising it by 10 % (to 3.3) moves the result by +0 %, from 4×10−4 to 4×10−4 m.
Modulus of member 1
The modulus of member 1, in Pa. A stiffer member stretches less, in inverse proportion. Raising it by 10 % (to 2.2×1011) moves the result by −6.82 %, from 4×10−4 to 3.727×10−4 m.
Modulus of member 2
The modulus of member 2, in Pa. A stiffer member stretches less, in inverse proportion. Raising it by 10 % (to 2.2×1011) moves the result by −2.27 %, from 4×10−4 to 3.909×10−4 m.
Modulus of member 3
The modulus of member 3, in Pa. A stiffer member stretches less, in inverse proportion. Raising it by 10 % (to 2.2×1011) moves the result by +0 %, from 4×10−4 to 4×10−4 m.
Area of member 1
The cross-section area of member 1, in m². A larger area stretches less, in inverse proportion. Raising it by 10 % (to 0.0011) moves the result by −6.82 %, from 4×10−4 to 3.727×10−4 m.
Area of member 2
The cross-section area of member 2, in m². A larger area stretches less, in inverse proportion. Raising it by 10 % (to 0.0011) moves the result by −2.27 %, from 4×10−4 to 3.909×10−4 m.
Area of member 3
The cross-section area of member 3, in m². A larger area stretches less, in inverse proportion. Raising it by 10 % (to 0.0011) moves the result by +0 %, from 4×10−4 to 4×10−4 m.
Worked example with the values above
- Member terms N·n·L/(EA): member 1 = 20,000 × 1 × 3 / (2×1011 × 0.001) = 3×10−4; member 2 = −10,000 × −0.5 × 4 / (2×1011 × 0.001) = 1×10−4 m.
- Sum of the terms: 4×10−4 m.
- δ = 4×10−4 m.
Assumptions and limits
- Pin joints, loads at the joints, small displacements and linear elastic members.
- Tension positive in both the real and virtual force systems.
Common mistakes
- Using compression positive in one system and tension positive in the other.
- Mixing the real force and the virtual force between members.
- Forgetting that unused members must have zero real or virtual force.
Statistics and uncertainty
Truss displacements are exact for pin-jointed bars with axial deformation only. Real joints are slightly flexible and members have length tolerances, so measured displacements are usually 5 % to 20 % higher than the pin-jointed prediction.
Where it is used
- Deflection and stiffness checks for trusses, braces and frames.
- Estimating thermal movement and strain energy of members.
Maths you need
- Virtual work: external virtual work equals internal virtual work
- Axial deformation: ΔL = NL/EA
- Strain energy: the work stored by elastic deformation